Puzzle Time!photo credit: Toni Blay via photopin cc |

In a "Lights Out" puzzle, the goal is to turn everything off, by hitting switches that toggle certain combinations on the board.

For instance, I might encounter a door guarded by laser beams. There are 4 laser beams bouncing around and if they are all turned off, I can make it through the door safely. If even one of them is on, I will get disintegrated. We'll call these lasers 1, 2, 3, and 4.

O O O O

In front of the laser-guarded door are 5 levers, which we'll call A, B, C, D, and E. Each lever can toggle one or more of the laser beams. So maybe if I hit A, it toggles 1 and 2 (turning them both off):

x x O O

*(x means off, O means on)*

but if I then hit B and it toggles 2 and 3, then 2 gets turned back on while 3 gets turned off:

x O x O

So, the puzzle is to find the right sequence of levers that toggles all of the laser beams into an "off" position.

This is made more fun/crazy/insane/mean if the GM applies a time limit, like a crushing wall moving in from the opposite side of the room.

### How to build it

There are probably a lot of methods, but simplicity for the GM is key from my perspective. So, I'll set up something like this and then arbitrarily select a solution. Let's say in the above example, I arbitrarily choose a solution as A, D, E.A few things about this: First, in puzzles like this, hitting a lever twice is equivalent to not hitting it at all (since it exactly inverses itself). Second, order is irrelevant, as it's all a binary system.

Third, your solution might not be unique - you could accidentally build in multiple solutions. Who cares - as long as the players can solve it (and we've already guaranteed that by starting with the solution and working backwards), then the players will be happy they found "the" solution. Just smile like you knew it all along.

OK. Back to the design. Each laser (1-4) must appear an

*odd*number of times in the solution. Once means it turns

**off**, while thrice means it goes off-on-

**off**, but twice would mean off-

**on**and thus be invalid.

So, fill in the chosen solution such that each laser appears an odd number of times:

**A: 12**

B:

C:

**D: 14**

**E: 13**

Hitting A, D, and E in any order will turn off all the lasers. Now we fill some junk into B and C however we want:

**A: 12**

B: 23

C: 24

**D: 14**

**E: 13**

Now the players can screw around with false solutions involving B and C. But check it out - I accidentally made some easier solutions. C and E works right away. So does B and D. We could either leave those in, or modify it, say by adding something:

**A: 12**

B: 234

C: 124

**D: 14**

**E: 13**

That makes it a bit different. Remember in all puzzles that it gets a lot harder once the players look at it, so even if there is an "obvious" solution once you see the key like this, it isn't necessarily easy for players.

### The next level

What if there are four lasers, as above, but one of them starts in the "off" position? Let's say laser 2 starts off and the other three are on.O x O O

Well, that means that laser two now needs to appear an

*even*number of times in the solution, and the others still need to appear an odd number of times. Same 5 levers, this time let's say A B D is the solution.

**A: 12**

**B: 13**

C: 1234

**D: 124**

E: 134

I filled in C and E with fluff. In the bolded lines (A/B/D), 1 appears thrice (odd), 2 appears twice (even), 3 appears once (odd), and 4 appears once (odd). Mission accomplished.

### Hints

Want hints for the players? Say, off a Mental check (INT in D&D)? Easy enough. If they succeed on your sinister and arbitrary check difficulty, then you can tell them things like:- Two of the levers aren't needed for the solution.
- Lever E isn't needed for the solution.

Or something similar.

Remember, the goal with puzzles is always to make it easy enough that players can solve it, but hard enough that they feel satisfaction from solving it!

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